Is it cheaper to boil water on high gas heat or low gas heat? Assume my time value is \$0. High gas heat adds more energy per second to the water but the pot may absorb the heat less efficiently, or maybe the shape of the flame is less good, or for other reasons diminishing returns are encountered. Low enough gas heat might never boil the water. Is there an efficient optimum in the middle that uses the least total gas to get the water to 100℃?
More generally: what are the factors in a real-life situation of warming something up that differ from the simplest model of a knob that makes "temperature" rise (characterises the entire fluid by one scalar), forgetting that something physical has to heat up something else physical?
 A: Here are some factors to consider:


*

*Thermal conduction can be modeled in a material by Fourier's law which states that
$$
  \mathbf q = -k \nabla T
$$
where $\mathbf q$ is local heat flux, $k$ is the thermal conductivity of the material, and $T$ is temperature.

*One aspect of this that can complicate real-world applications is anisotropy or inhomogeneity of conductivity ($k$ may depend on direction and/or position in a material, or you could have two materials with different conductivities), and also the fact that conductivity can depend on temperature.  These can make Fourier's law a difficult PDE to work with.  In light of this, the pot, the air, and the water all have different thermal conductivities and these may depend on temperature (although I'm by no means familiar with the thermal conductivities of different materials).
There are probably others that don't immediately come to mind.
Cheers!
P.S. You might find this helpful/interesting Heat transfer between two surfaces
A: It is almost certainly cheaper to use high heat for the following reasons:


*

*A hotter-than-room-temperature pot will necessarily be losing heat to its surroundings by emitting approximately blackbody radiation. The flux (power per unit area) emitted is $\sigma T_\text{pot}^4$. There is also a $\sigma T_\text{room}^4$ heating of the pot from the surroundings.

*A hot pot will also conduct heat to the air, which will transport this away via convection. In an ideal situation, this is described by Newton's law of cooling, which states that the flux is $h(T_\text{pot} - T_\text{room})$ for some constant $h$.

*If the pot has an area $A$, the heat lost during the whole process is
\begin{align}
Q_\text{lost} & = A \int_{t_\text{begin}}^{t_\text{end}} \left(\sigma \left(T_\text{pot}^4 - T_\text{room}^4\right) + h \left(T_\text{pot} - T_\text{room}\right)\right) \mathrm{d}t \\
& = A \int_{T_\text{room}}^{T_\text{boil}} \left(\sigma \left(T_\text{pot}^4 - T_\text{room}^4\right) + h \left(T_\text{pot} - T_\text{room}\right)\right) \left(\frac{\mathrm{d}T_\text{pot}}{\mathrm{d}t}\right)^{-1} \mathrm{d}T_\text{pot}.
\end{align}
The integrand in the second line is clearly a monotonically decreasing function of the heating rate.

*Using higher heat will therefore lead to lower losses from the pot along the way.


The only worry is that a greater fraction of heat from the fire will be lost directly to the environment when the gas is turned up, but I cannot see any mechanism for accomplishing this to a significant degree. The fraction of the fire's heat going into the pot through radiative transfer is fixed - it is just the cross section of the pot. Furthermore, when heating a pot of water gently, I anecdotally feel more heat coming off the warm pot than coming directly from the fire.
